If `y=sqrt(x)^(sqrt(x)^(sqrt(x)^sqrt(x)...00)), "then the value of"(dy)/(dx)"is"-`

Options

A. `(xy^(2))/(2-ylogx)`
B. `(x^(2))/(y(2-ylogx))`
C. `(y^(2))/(x(2-ylogx))`
D. `(y^(2))/(x(2+ylogx))`

Answer

To solve the problem where \( y = \sqrt{x}^{\sqrt{x}^{\sqrt{x}^{\cdots}}} \), we will follow these steps:

### Step 1: Define the equation
We start by rewriting the equation in a more manageable form. Since the expression is self-referential, we can express it as:
\[ y = \sqrt{x}^y \]

### Step 2: Take the logarithm of both sides
Next, we take the natural logarithm of both sides to simplify the exponent:
\[ \ln(y) = y \ln(\sqrt{x}) \]
We know that \( \ln(\sqrt{x}) = \frac{1}{2} \ln(x) \), so we can rewrite the equation as:
\[ \ln(y) = y \cdot \frac{1}{2} \ln(x) \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \). Using implicit differentiation on the left side and the product rule on the right side, we get:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \ln(x) \frac{dy}{dx} + y \cdot \frac{1}{2} \cdot \frac{1}{x} \]

### Step 4: Rearranging the equation
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{1}{y} \frac{dy}{dx} - \frac{1}{2} \ln(x) \frac{dy}{dx} = \frac{y}{2x} \]
Factoring out \( \frac{dy}{dx} \) from the left side gives us:
\[ \left( \frac{1}{y} - \frac{1}{2} \ln(x) \right) \frac{dy}{dx} = \frac{y}{2x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\frac{y}{2x}}{\frac{1}{y} - \frac{1}{2} \ln(x)} \]
Multiplying the numerator and denominator by \( y \) gives us:
\[ \frac{dy}{dx} = \frac{y^2}{2x \left( 1 - \frac{1}{2} y \ln(x) \right)} \]

### Final Answer
Thus, the value of \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{y^2}{2x \left( 1 - \frac{1}{2} y \ln(x) \right)} \]